How do you find the limit of #tan^-1(1/x)# as #x->0^+#? Calculus Limits Determining Limits Algebraically 1 Answer Andrea S. Feb 8, 2017 #lim_(x->0^+) arctan(1/x) = pi/2# Explanation: #lim_(x->0^+) arctan(1/x)# substitute: #y= 1/x#, with #lim_(x->0^+) y(x) = +oo# So: #lim_(x->0^+) arctan(1/x) = lim_(y->+oo) arctan(y) = pi/2# Answer link Related questions How do you find the limit #lim_(x->5)(x^2-6x+5)/(x^2-25)# ? How do you find the limit #lim_(x->3^+)|3-x|/(x^2-2x-3)# ? How do you find the limit #lim_(x->4)(x^3-64)/(x^2-8x+16)# ? How do you find the limit #lim_(x->2)(x^2+x-6)/(x-2)# ? How do you find the limit #lim_(x->-4)(x^2+5x+4)/(x^2+3x-4)# ? How do you find the limit #lim_(t->-3)(t^2-9)/(2t^2+7t+3)# ? How do you find the limit #lim_(h->0)((4+h)^2-16)/h# ? How do you find the limit #lim_(h->0)((2+h)^3-8)/h# ? How do you find the limit #lim_(x->9)(9-x)/(3-sqrt(x))# ? How do you find the limit #lim_(h->0)(sqrt(1+h)-1)/h# ? See all questions in Determining Limits Algebraically Impact of this question 30958 views around the world You can reuse this answer Creative Commons License